Largest int from concatenated ints

Largest int from concatenated ints
You are encouraged to solve this task according to the task description, using any language you may know.

Task

Given a set of positive integers, write a function to order the integers in such a way that the concatenation of the numbers forms the largest possible integer and return this integer.

Use the following two sets of integers as tests   and   show your program output here.

•   {1, 34, 3, 98, 9, 76, 45, 4}
•   {54, 546, 548, 60}

Possible algorithms

1. A solution could be found by trying all combinations and return the best.
2. Another way to solve this is to note that in the best arrangement, for any two adjacent original integers X and Y, the concatenation X followed by Y will be numerically greater than or equal to the concatenation Y followed by X.
3. Yet another way to solve this is to pad the integers to the same size by repeating the digits then sort using these repeated integers as a sort key.

See also

•   Algorithms: What is the most efficient way to arrange the given numbers to form the biggest number?
•   Constructing the largest number possible by rearranging a list

11l

Translation of: Python

F maxnum(x)
   V maxlen = String(max(x)).len
   R sorted(x.map(v -> String(v)), key' i -> i * (@maxlen * 2 I/ i.len), reverse' 1B).join(‘’)

L(numbers) [[212, 21221], [1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]]
   print("Numbers: #.\n  Largest integer: #15".format(numbers, maxnum(numbers)))

Output:

Numbers: [212, 21221]
  Largest integer:        21221221
Numbers: [1, 34, 3, 98, 9, 76, 45, 4]
  Largest integer:    998764543431
Numbers: [54, 546, 548, 60]
  Largest integer:      6054854654

Ada

The algorithmic idea is to apply a twisted comparison function:

function Order(Left, Right: Natural) return Boolean is
      ( (Img(Left) & Img(Right)) > (Img(Right) & Img(Left)) );

This function converts the parameters Left and Right to strings and returns True if (Left before Right) exceeds (Right before Left). It needs Ada 2012 -- the code for older versions of Ada would be more verbose.

The rest is straightforward: Run your favourite sorting subprogram that allows to use the function "Order" instead of standard comparison operators ("<" or ">" or so) and print the results:

with Ada.Text_IO, Ada.Containers.Generic_Array_Sort;

procedure Largest_Int_From_List is
   
   function Img(N: Natural) return String is
      S: String := Integer'Image(N);
   begin
      return S(S'First+1 .. S'Last); -- First character is ' '
   end Img;
   
   function Order(Left, Right: Natural) return Boolean is
      ( (Img(Left) & Img(Right)) > (Img(Right) & Img(Left)) );
   
   type Arr_T is array(Positive range <>) of Natural;
   
   procedure Sort is new Ada.Containers.Generic_Array_Sort
     (Positive, Natural, Arr_T, Order);
   
   procedure Print_Sorted(A: Arr_T) is
      B: Arr_T := A;
   begin
      Sort(B);
      for Number of B loop
	 Ada.Text_IO.Put(Img(Number));
      end loop;
      Ada.Text_IO.New_Line;
   end Print_Sorted;
   
begin
   Print_Sorted((1, 34, 3, 98, 9, 76, 45, 4));
   Print_Sorted((54, 546, 548, 60));
end Largest_Int_From_List;

Aime

largest(...)
{
    index x;
    for (, integer e in xcall(list).__list) {
        x[999999999 - 9.times(b_, data(), e).b_size(9).atoi] = e;
    }
    x.ucall(o_, 0);
    o_newline();
}

main(void)
{
    largest(1, 34, 3, 98, 9, 76, 45, 4);
    largest(54, 546, 548, 60);
    0;
}

works for input up to 999999999.

Output:

998764543431
6054854654

ALGOL 68

Using method 2 - first sorting into first digit order and then comparing concatenated pairs.

BEGIN
    # returns the integer value of s #
    OP TOINT = ( STRING s)INT:
    BEGIN
        INT result := 0;
        FOR s pos FROM LWB s TO UPB s DO
            result *:= 10 +:= ( ABS s[ s pos ] - ABS "0" )
        OD;
        result
    END # TOINT # ;
    # returns the first digit of n #
    OP FIRSTDIGIT = ( INT n )INT:
    BEGIN
        INT result := ABS n;
        WHILE result > 9 DO result OVERAB 10 OD;
        result
    END # FIRSTDIGIT # ;
    # returns a string representaton of n #
    OP TOSTRING = ( INT n )STRING: whole( n, 0 );
    # returns an array containing the values of a sorted such that concatenating the values would result in the largest value #
    OP CONCATSORT = ( []INT a )[]INT:
       IF LWB a >= UPB a THEN
           # 0 or 1 element(s) #
           a
       ELSE
           # 2 or more elements #
           [ 1 : ( UPB a - LWB a ) + 1 ]INT result := a[ AT 1 ];
           # sort the numbers into reverse first digit order #
           FOR o pos FROM UPB result - 1 BY -1 TO 1
           WHILE BOOL swapped := FALSE;
                 FOR i pos TO o pos DO
                     IF FIRSTDIGIT result[ i pos ] < FIRSTDIGIT result[ i pos + 1 ] THEN
                         INT t = result[ i pos + 1 ];
                         result[ i pos + 1 ] := result[ i pos ];
                         result[ i pos     ] := t;
                         swapped             := TRUE
                      FI
                 OD;
                 swapped
           DO SKIP OD;
           # now re-order adjacent numbers so they have the highest concatenated value #
           WHILE BOOL swapped := FALSE;
                 FOR i pos TO UPB result - 1 DO
                     STRING l := TOSTRING result[ i pos     ];
                     STRING r := TOSTRING result[ i pos + 1 ];
                     IF TOINT ( l + r ) < TOINT ( r + l ) THEN
                         INT t = result[ i pos + 1 ];
                         result[ i pos + 1 ] := result[ i pos ];
                         result[ i pos     ] := t;
                         swapped             := TRUE
                     FI
                 OD;
                 swapped
           DO SKIP OD;
           result
       FI # CONCATSORT # ;
    # prints the array a #
    OP PRINT = ( []INT a )VOID:
       FOR a pos FROM LWB a TO UPB a DO
            print( ( TOSTRING a[ a pos ] ) )
       OD # PRINT # ;

    # task test cases #
    PRINT CONCATSORT []INT( 1, 34, 3, 98, 9, 76, 45, 4 );
    print( ( newline ) );
    PRINT CONCATSORT []INT( 54, 546, 548, 60 );
    print( ( newline ) )

END

Output:

998764543431
6054854654

Arturo

largestConcInt: function [arr]->
    max map permutate arr 's [
        to :integer join map s => [to :string]
    ]

loop [[1 34 3 98 9 76 45 4] [54 546 548 60]] 'a ->
    print largestConcInt a

AutoHotkey

LargestConcatenatedInts(var){
	StringReplace, var, A_LoopField,%A_Space%,, all
	Sort, var, D`, fConcSort
	StringReplace, var, var, `,,, all
	return var
}

ConcSort(a, b){
	m := a . b	, n := b . a
    return m < n ? 1 : m > n ? -1 : 0
}

Examples:

d =
(
1, 34, 3, 98, 9, 76, 45, 4
54, 546, 548, 60
4 , 45, 54, 5
)
loop, parse, d, `n
	MsgBox % LargestConcatenatedInts(A_LoopField)

Output:

998764543431
6054854654
554454

AWK

Works with: gawk version 4.0

function cmp(i1, v1, i2, v2, u1, u2) {
	u1 = v1""v2;
	u2 = v2""v1;
        return (u2 - u1)
}
function largest_int_from_concatenated_ints(X) {
 	PROCINFO["sorted_in"]="cmp";
	u="";
	for (i in X) u=u""X[i];
	return u
} 

BEGIN {
	split("1 34 3 98 9 76 45 4",X);
	print largest_int_from_concatenated_ints(X)

	split("54 546 548 60",X);
	print largest_int_from_concatenated_ints(X)
}

Output:

998764543431
6054854654

BBC BASIC

      DIM Nums%(10)
      Nums%()=1,34,3,98,9,76,45,4
      PRINT FNlargestint(8)
      Nums%()=54,546,548,60
      PRINT FNlargestint(4)
      END
      
      DEF FNlargestint(len%)
      LOCAL i%,l$,a$,b$,sorted%
      REPEAT
        sorted%=TRUE
        FOR i%=0 TO len%-2
          a$=STR$Nums%(i%)
          b$=STR$Nums%(i%+1)
          IF a$+b$<b$+a$ SWAP Nums%(i%),Nums%(i%+1):sorted%=FALSE
        NEXT
      UNTIL sorted%
      FOR i%=0 TO len%-1
        l$+=STR$Nums%(i%)
      NEXT
      =l$

Output:

998764543431
6054854654

Bracmat

( ( maxnum
  =   A Z F C
    .   !arg:#
      |   !arg
        :   %@?F
            ?
            ( #%@?C
            & ( str$(!F !C)+-1*str$(!C !F):~<0
              | !C:?F
              )
            & ~
            )
            ?
      | !arg:?A !F ?Z&!F maxnum$(!A !Z)
  )
& out$(str$(maxnum$(1 34 3 98 9 76 45 4)))
& out$(str$(maxnum$(54 546 548 60)))
);

Output:

998764543431
6054854654

C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int catcmp(const void *a, const void *b)
{
	char ab[32], ba[32];
	sprintf(ab, "%d%d", *(int*)a, *(int*)b);
	sprintf(ba, "%d%d", *(int*)b, *(int*)a);
	return strcmp(ba, ab);
}

void maxcat(int *a, int len)
{
	int i;
	qsort(a, len, sizeof(int), catcmp);
	for (i = 0; i < len; i++)
		printf("%d", a[i]);
	putchar('\n');
}

int main(void)
{
	int x[] = {1, 34, 3, 98, 9, 76, 45, 4};
	int y[] = {54, 546, 548, 60};

	maxcat(x, sizeof(x)/sizeof(x[0]));
	maxcat(y, sizeof(y)/sizeof(y[0]));

	return 0;
}

Output:

998764543431
6054854654

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var source1 = new int[] { 1, 34, 3, 98, 9, 76, 45, 4 };
        var source2 = new int[] { 54, 546, 548, 60 };

        var largest1 = LargestPossibleSequence(source1);
        var largest2 = LargestPossibleSequence(source2);

        Console.WriteLine("The largest possible integer from set 1 is: {0}", largest1);
        Console.WriteLine("The largest possible integer from set 2 is: {0}", largest2);
    }

    static long LargestPossibleSequence(int[] ints)
    {
        return long.Parse(string.Join("", ints.OrderBy(i => i, new IntConcatenationComparer()).Reverse()));
    }
}

class IntConcatenationComparer : IComparer<int>
{
    public int Compare(int x, int y)
    {
        var xy = int.Parse(x.ToString() + y.ToString());
        var yx = int.Parse(y.ToString() + x.ToString());

        return xy - yx;
    }
}

Output:

The largest possible integer from set 1 is: 998764543431
The largest possible integer from set 2 is: 6054854654

C++

#include <iostream>
#include <sstream>
#include <algorithm>
#include <vector>
#include <string>

std::string findLargestConcat ( std::vector< int > & mynumbers ) {
   std::vector<std::string> concatnumbers ;
   std::sort ( mynumbers.begin( ) , mynumbers.end( ) ) ;
   do {
      std::ostringstream numberstream ;
      for ( int i : mynumbers ) 
	 numberstream << i ;
      concatnumbers.push_back( numberstream.str( ) ) ;
   } while ( std::next_permutation( mynumbers.begin( ) ,
	    mynumbers.end( ) )) ;
   return *( std::max_element( concatnumbers.begin( ) ,
	 concatnumbers.end( ) ) ) ;
}

int main( ) {
   std::vector<int> mynumbers = { 98, 76 , 45 , 34, 9 , 4 , 3 , 1 } ;
   std::vector<int> othernumbers = { 54 , 546 , 548 , 60 } ;
   std::cout << "The largest concatenated int is " <<
      findLargestConcat( mynumbers ) << " !\n" ;
   std::cout << "And here it is " << findLargestConcat( othernumbers ) 
      << " !\n" ;
   return 0 ;
}

Output:

The largest concatenated int is 998764543431 !
And here it is 6054854654 !

Ceylon

Translation of: Kotlin

Works with: Ceylon version 1.3.3

shared void run() {
	
	function comparator(Integer x, Integer y) {
		assert (is Integer xy = Integer.parse(x.string + y.string),
			is Integer yx = Integer.parse(y.string + x.string));
		return yx <=> xy;
	}
	
	function biggestConcatenation({Integer*} ints) => "".join(ints.sort(comparator));
	
	value test1 = {1, 34, 3, 98, 9, 76, 45, 4};
	print(biggestConcatenation(test1));
	
	value test2 = {54, 546, 548, 60};
	print(biggestConcatenation(test2));
}

Clojure

(defn maxcat [coll]
  (read-string
    (apply str
           (sort (fn [x y]
                   (apply compare
                          (map read-string [(str y x) (str x y)])))
                 coll))))

(prn (map maxcat [[1 34 3 98 9 76 45 4] [54 546 548 60]]))

Output:

(998764543431 6054854654)

Common Lisp

Sort by two-by-two comparison of largest concatenated result

(defun int-concat (ints)
  (read-from-string (format nil "~{~a~}" ints)))

(defun by-biggest-result (first second)
  (> (int-concat  (list first second)) (int-concat (list second first))))

(defun make-largest-int (ints) 
  (int-concat (sort ints #'by-biggest-result)))

Output:

> (make-largest-int '(1 34 3 98 9 76 45 4))
998764543431

> (make-largest-int '(54 546 548 60))
6054854654

Variation around the sort with padded most significant digit

;; Sort criteria is by most significant digit with least digits used as a tie
;; breaker

(defun largest-msd-with-less-digits (x y)
  (flet ((first-digit (x)
           (digit-char-p (aref x 0))))
    (cond ((> (first-digit x)
              (first-digit y))
           t)
          ((> (first-digit y)
              (first-digit x))
           nil)
          ((and (= (first-digit x)
                   (first-digit y))
                (> (length x)
                   (length y)))
           nil)
          (t t))))

(loop
  :for input :in '((54 546 548 60) (1 34 3 98 9 76 45 4))
  :do (format t "~{~A~}~%"
              (sort (mapcar #'write-to-string input)
                    #'largest-msd-with-less-digits)))

Output:

6054548546
998764453341

D

The three algorithms. Uses the second module from the Permutations Task.

import std.stdio, std.algorithm, std.conv, std.array, permutations2;

auto maxCat1(in int[] arr) pure @safe {
    return arr.to!(string[]).permutations.map!join.reduce!max;
}

auto maxCat2(in int[] arr) pure nothrow @safe {
    return arr.to!(string[]).sort!q{b ~ a < a ~ b}.join;
}

auto maxCat3(in int[] arr) /*pure nothrow @safe*/ {
    immutable maxL = arr.reduce!max.text.length;
    return arr.to!(string[])
           .schwartzSort!(s => s.replicate(maxL/s.length + 1), "a > b")
           .join;
}

void main() {
    const lists = [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]];
    [&maxCat1, &maxCat2, &maxCat3].map!(cat => lists.map!cat).writeln;
}

Output:

[["998764543431", "6054854654"], ["998764543431", "6054854654"], ["998764543431", "6054854654"]]

Delphi

See Pascal.

EasyLang

func con a b .
   t = 10
   while b >= t
      t *= 10
   .
   return a * t + b
.
func$ max a[] .
   n = len a[]
   for i to n - 1
      for j = i + 1 to n
         if con a[i] a[j] < con a[j] a[i]
            swap a[i] a[j]
         .
      .
   .
   for v in a[]
      r$ &= v
   .
   return r$
.
print max [ 1 34 3 98 9 76 45 4 ]
print max [ 54 546 548 60 ]

Output:

998764543431
6054854654

Elixir

defmodule RC do
  def largest_int(list) do
    sorted = Enum.sort(list, fn x,y -> "#{x}#{y}" >= "#{y}#{x}" end)
    Enum.join(sorted)
  end
end

IO.inspect RC.largest_int [1, 34, 3, 98, 9, 76, 45, 4]
IO.inspect RC.largest_int [54, 546, 548, 60]

Output:

"998764543431"
"6054854654"

Erlang

-module( largest_int_from_concatenated ).

-export( [ints/1, task/0] ).

ints( Ints ) ->
	Int_strings = [erlang:integer_to_list(X) || X <- Ints],
	Pad_ints = [{X ++ X, X} || X <- Int_strings],
	erlang:list_to_integer( lists:append([Int || {_Pad, Int} <- lists:reverse(lists:sort(Pad_ints))]) ).

task() ->
	[io:fwrite("Largest ~p from ~p~n", [ints(X), X]) || X <- [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]]].

Output:

8> largest_int_from_concatenated:task().
Largest 998764543431 from [1,34,3,98,9,76,45,4]
Largest 6054854654 from [54,546,548,60]

F#

// Form largest integer which is a permutation from a list of integers. Nigel Galloway: March 21st., 2018
let fN g = List.map (string) g |> List.sortWith(fun n g->if n+g<g+n then 1 else -1) |> System.String.Concat

Output:

fN [1; 34; 3; 98; 9; 76; 45; 4] -> "998764543431"
fN [54; 546; 548; 60]           -> "6054854654"

Factor

Using algorithm 3:

USING: assocs io kernel math qw sequences sorting ;
IN: rosetta-code.largest-int

: pad ( target seq -- padded )
    2dup length / swap <repetition> concat swap head ;
    
: largest-int ( seq -- )
    dup dup [ length ] map supremum    ! find longest length so we know how much to pad
    [ swap pad ] curry map             ! pad the integers
    <enum> sort-values                 ! sort the padded integers
    keys                               ! find the original indices of the sorted integers
    swap nths                          ! order non-padded integers according to their sorted order
    reverse concat print ;             
    
qw{ 1 34 3 98 9 76 45 4 } qw{ 54 546 548 60 } [ largest-int ] bi@

Output:

998764543431
6054854654

Or alternatively, a translation of F#.

Translation of: F#

USING: kernel math.order qw sequences sorting ;

: fn ( seq -- str )
    [ 2dup swap [ append ] 2bi@ after? +lt+ +gt+ ? ] sort concat ;

Output:

qw{ 1 34 3 98 9 76 45 4 } qw{ 54 546 548 60 } [ fn ] bi@

--- Data stack:
"998764543431"
"6054854654"

Fortran

There is often a potential ambiguity when reading numbers. While three definitely names the Platonic number notion, 3 might instead be regarded as being a text that happens to have the glyph of a number but is not a number. This sort of discussion arises when a spreadsheet has read in a text file and behold! numbers are on the display and they look just like what is displayed when numbers are being shown, but, they are not numbers, they are only drawn that way. Within the spreadsheet they are parts of some text, and the notion that takes over is one of a "blunt, heavy object", not alas close to hand.

So, the plan is to regard the numbers as being text sequences aligned to the left, containing only digit characters of course - except for the fact that CHARACTER variables often end up having trailing spaces. F2003 formalised a scheme whereby such variables can be "cut-to-fit" as execution proceeds but with earlier Fortrans the standard method is to pay attention to the number of characters in use. F90 introduced a function LEN_TRIM(text) to return the index of the last non-blank character in a text so the only problem now is to decide on how long might the largest number be (and by representing numbers as text strings, there is no difficulty with the limits of INTEGER*2 or INTEGER*4 etc.), and what will be the maximum number of numbers. By devising a subroutine to do the work, these issues can be handled by the caller that is providing the data. The subroutine however intends to sort the collection of texts. This could be done by damaging its parameter which might be regarded as impolite or even unwanted so instead the sort is effected via an array XLAT and juggling its values. This has the advantage that the possibly large elements of the text array are not being moved about, but means that the subroutine must be able to have an XLAT array that is "large enough". F90 standardised the ability for a routine to declare such an array at run-time; previously, arrays within a subroutine (or indeed anywhere) had to have a size fixed at compilation time. In the past this might have been handled by the caller supplying such an array as an additional parameter.

Passing arrays as parameters can be tricky, especially for multi-dimensional arrays. This uses the old style whereby the size is left unstated via the * in TEXT(*), though one could use TEXT(N) instead - but at the risk that the actual value of N is wrong and array index checking might be confused thereby. Still earlier one would simply place some integer value there, any valid integer, as in TEXT(666), and not worry about bound checking at all because old-style compilers did not produce checking code even if it was wanted. F90 standardised the MODULE protocol, within which the size is specified as TEXT(:) whereby secret additional parameters are supplied that contain the actual bound information and bound checking will be correct, possibly not so if the TEXT(N) form is used instead and N is wrong. This extra overhead in every use is possibly better than undetected errors in some uses...

The sorting of the text array was to be by the notorious BubbleSort, taking advantage of the fact that each pass delivers the maximum value of the unsorted portion to its final position: the output could thereby be produced as the sort worked. Rather than mess about with early termination (no element being swapped) or attention to the bounds within which swapping took place, attention concentrated upon the comparison. Because of the left-alignment of the texts, a simple comparison seemed sufficient until I thought of unequal text lengths and then the following example. Suppose there are two numbers, 5, and one of 54, 55, or 56 as the other. Via normal comparisons, the 5 would always be first (because short texts are considered expanded with trailing spaces when compared against longer texts, and a space precedes every digit) however the biggest ordering is 5 54 for the first case but 56 5 for the last. This possibility is not exemplified in the specified trial sets. So, a more complex comparison is required. One could of course write a suitable function and consider the issue there but instead the comparison forms the compound text in the same manner as the result will be, in the two ways AB and BA, and looks to see which yields the bigger sequence. This need only be done for unequal length text pairs.

The source is F77 style, except for the declaration of XLAT(N), the use of <N> in the FORMAT statements instead of some large constant or similar, and the ability to declare an array via constants as in (/"5","54"/) rather than mess about declaring arrays and initialising them separately. The I0 format code to convert a number (an actual number) into a digit string aligned leftwards in a CHARACTER variable of sufficient size is also a F90 introduction, though the B6700 compiler allowed a code J instead. This last is to demonstrate usage of actual numbers for those unpersuaded by the argument for ambiguity that allows for texts. If the I0 format code is unavailable then I9 (or some suitable size) could be used, followed by text = ADJUSTL(text), except that this became an intrinsic function only in F90, so perhaps you will have to write a simple alignment routine.

      SUBROUTINE SWAP(A,B)	!Why can't the compiler supply these!
       INTEGER A,B,T
        T = B
        B = A
        A = T
      END

      SUBROUTINE BIGUP(TEXT,N)	!Outputs the numbers in TEXT to give the biggest number.
       CHARACTER*(*) TEXT(*)	!The numbers as text, aligned left.
       INTEGER N		!The number of them.
       INTEGER XLAT(N),L(N)	!An index and a set of lengths.
       INTEGER I,J,M		!Assorted steppers.
       INTEGER TI,TJ		!Fingers to a text.
       INTEGER LI,LJ		!Lengths of the fingered texts.
       INTEGER MSG		!I/O unit number.
       COMMON /IODEV/ MSG	!Old style.
        DO I = 1,N	!Step through my supply of texts.
          XLAT(I) = I		!Preparing a finger to them.
          L(I) = LEN_TRIM(TEXT(I))	!And noting their last non-blank.
        END DO		!On to the next.
        WRITE (MSG,1) "Supplied",(TEXT(I)(1:L(I)), I = 1,N)	!Show the grist.
    1   FORMAT (A12,":",<N>(A,","))	!Instead of <N>, 666 might suffice.
Crude bubblesort. No attempt at noting the bounds of swaps made.
        DO M = N,1,-1	!Just for fun, go backwards.
          DO I = 2,M		!Start a scan.
            J = I - 1		!Comparing element I to element I - 1.
            TI = XLAT(I)	!Thus finger the I'th text in XLAT order.
            TJ = XLAT(J)	!And its supposed predecessor.
            LI = L(TI)		!The length of the fingered text.
            LJ = L(TJ)		!All this to save on typing below.
            IF (LI .EQ. LJ) THEN	!If the texts are equal lengths,
              IF (TEXT(TI).LT.TEXT(TJ)) CALL SWAP(XLAT(I),XLAT(J))	!A simple comparison.
             ELSE	!But if not, construct the actual candidate texts for comparison.
              IF (TEXT(TI)(1:LI)//TEXT(TJ)(1:LJ)	!These two will be the same length.
     1        .LT.TEXT(TJ)(1:LJ)//TEXT(TI)(1:LI))	!Just as above.
     2        CALL SWAP(XLAT(I),XLAT(J))	!J shall now follow I.
            END IF			!So much for that comparison.
          END DO		!On to the next.
        END DO	!The original plan was to reveal element XLAT(M) as found.
        WRITE (MSG,2) "Biggest",(TEXT(XLAT(I))(1:L(XLAT(I))),I = N,1,-1)	!But, all at once is good too.
    2   FORMAT (A12,":",<N>(A," "))	!The space maintains identity.
      END	!That was fun.

      PROGRAM POKE
      CHARACTER*4 T1(10)	!Prepare some example arrays.
      CHARACTER*4 T2(4)		!To hold the specified examples.
      INTEGER MSG
      COMMON /IODEV/ MSG
      DATA T1(1:8)/"1","34","3","98","9","76","45","4"/
      DATA T2/"54","546","548","60"/
      MSG = 6		!Standard output.
      WRITE (MSG,1)
    1 FORMAT ("Takes a list of integers and concatenates them so as ",
     1 "to produce the biggest possible number.",/,
     2 "The result is shown with spaces between the parts ",
     3 "to show provenance. Ignore them otherwise."/)
      CALL BIGUP(T1,8)

      WRITE (MSG,*)
      CALL BIGUP(T2,4)

      WRITE (MSG,*) "These are supplied in lexicographical order..."
      CALL BIGUP((/"5","54"/),2)

      WRITE (MSG,*) "But this is not necessarily the biggest order."
      CALL BIGUP((/"5","56"/),2)

      WRITE (MSG,*) "And for those who count..."
      DO I = 1,10
        WRITE (T1(I),"(I0)") I	!This format code produces only the necessary text.
      END DO			!Thus, the numbers are aligned left in the text field.
      CALL BIGUP(T1,10)
      END

Output: the Fortran compiler ignores spaces when reading fortran source, so, hard-core fortranners should have no difficulty doing likewise for the output...

Takes a list of integers and concatenates them so as to produce the biggest possible number.
The result is shown with spaces between the parts to show provenance. Ignore them otherwise.

    Supplied:1,34,3,98,9,76,45,4,
     Biggest:9 98 76 45 4 34 3 1

    Supplied:54,546,548,60,
     Biggest:60 548 546 54
 These are supplied in lexicographical order...
    Supplied:5,54,
     Biggest:5 54
 But this is not necessarily the biggest order.
    Supplied:5,56,
     Biggest:56 5
 And for those who count...
    Supplied:1,2,3,4,5,6,7,8,9,10,
     Biggest:9 8 7 6 5 4 3 2 1 10

FreeBASIC

#define MAXDIGITS 8

function catint( a as string, b as string ) as uinteger
    return valint(a+b)
end function

function grt( a as string, b as string ) as boolean
    return catint(a, b)>catint(b, a)
end function

sub shellsort( a() as string )
    'quick and dirty shellsort, not the focus of this exercise
    dim as uinteger gap = ubound(a), i, j, n=ubound(a)
    dim as string temp
    do 
        gap = int(gap / 2.2)
        for i=gap to n
            temp = a(i)
            j=i
            while j>=gap andalso grt( a(j-gap), temp )
                a(j) = a(j - gap)
                j -= gap
            wend 
            a(j) = temp
        next i
    loop until gap = 1
end sub

sub sort_and_print( a() as string )
    dim as uinteger i
    dim as string outstring = ""
    shellsort(a())
    for i=0 to ubound(a)
        outstring = a(i)+outstring
    next i
    print outstring
end sub

dim as string set1(8) = {"1", "34", "3", "98", "9", "76", "45", "4"}
dim as string set2(4) = {"54", "546", "548", "60"}

sort_and_print(set1())
sort_and_print(set2())

Output:

998764543431
6054854654 

Frink

a = [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]]
f = {|p| parseInt[join["",p]] }
for s = a
   println[max[map[f, s.lexicographicPermute[]]]]

Output:

998764543431
6054854654

Gambas

Click this link to run this code

'Largest int from concatenated ints

Public Sub Main()
Dim iList1 As Integer[] = [1, 34, 3, 98, 9, 76, 45, 4]      'Integer list 1
Dim iList2 As Integer[] = [54, 546, 548, 60]                'Integer list 2

Calc(iList1)                                                'Send List 1 to Calc routine 
Calc(iList2)                                                'Send List 2 to Calc routine

End
'_________________________________________________________________________________________

Public Sub Calc(iList As Integer[])
Dim siCount1, siCount2, siCounter As Short                  'Counters
Dim sList As New String[]                                   'To hold converted integers
Dim bTrigger As Boolean                                     'To trigger a found match

For Each siCount1 In iList                                  'For each integer in the list..
  sList.Add(Str(siCount1))                                  'Convert to a string and add to sList
  If Len(Str(siCount1)) > siCounter Then                    'If the length of the string is greater than siCounter then..
    siCounter = Len(Str(siCount1))                          'siCounter = length of the string
  End If
Next

For siCount1 = 0 To sList.Max                               'For each item in sList
  If Len(sList[siCount1]) < siCounter Then                  'If the length of the string is less that siCounter then..
    sList[siCount1] &= Right(sList[siCount1], 1)            'Add the same digit to the string e.g. in list 1 "9" becomes "99", list 2 "54" becomes "544"
  End If
Next

sList.Sort(gb.Descent)                                      'Sort the list in decending order

For siCount1 = 0 To sList.Max                               'For each item in sList
  bTrigger = False                                          'Set bTrigger to False
  For siCount2 = 0 To iList.Max                             'Loop through each item in iList
    If Val(sList[siCount1]) = iList[siCount2] Then          'If the value of each is the same e.g. "98" = 98 then
      bTrigger = True                                       'Set bTrigger to True
      Continue                                              'Exit the loop
    Endif
  Next
  If Not bTrigger Then                                      'If there was no match e.g. there is no "99" then.. 
    sList[siCount1] = Left(sList[siCount1], siCounter - 1)  'Strip out the end digit e.g. "99" becomes 9 again
  End If
Next

Print Val(sList.Join(""))                                   'Join all items in sList together and print

End

Output:

998764543431
6054854654

Go

// Variation of method 3.  Repeat digits to at least the size of the longest,
// then sort as strings.
package main

import (
    "fmt"
    "math/big"
    "sort"
    "strconv"
    "strings"
)

type c struct {
    i     int
    s, rs string
}

type cc []*c

func (c cc) Len() int           { return len(c) }
func (c cc) Less(i, j int) bool { return c[j].rs < c[i].rs }
func (c cc) Swap(i, j int)      { c[i], c[j] = c[j], c[i] }

// Function required by task.  Takes a list of integers, returns big int.
func li(is ...int) *big.Int {
    ps := make(cc, len(is))
    ss := make([]c, len(is))
    ml := 0
    for j, i := range is {
        p := &ss[j]
        ps[j] = p
        p.i = i
        p.s = strconv.Itoa(i)
        if len(p.s) > ml {
            ml = len(p.s)
        }
    }
    for _, p := range ps {
        p.rs = strings.Repeat(p.s, (ml+len(p.s)-1)/len(p.s))
    }
    sort.Sort(ps)
    s := make([]string, len(ps))
    for i, p := range ps {
        s[i] = p.s
    }
    b, _ := new(big.Int).SetString(strings.Join(s, ""), 10)
    return b
}

func main() {
    fmt.Println(li(1, 34, 3, 98, 9, 76, 45, 4))
    fmt.Println(li(54, 546, 548, 60))
}

Output:

998764543431
6054854654

Groovy

def largestInt = { c -> c.sort { v2, v1 -> "$v1$v2" <=> "$v2$v1" }.join('') as BigInteger }

Testing:

assert largestInt([1, 34, 3, 98, 9, 76, 45, 4]) == 998764543431
assert largestInt([54, 546, 548, 60]) == 6054854654

Haskell

Compare repeated string method

import Data.List (sortBy)
import Data.Ord (comparing)

main = print (map maxcat [[1,34,3,98,9,76,45,4], [54,546,548,60]] :: [Integer])
    where
      sorted xs = let pad x  = concat $ replicate (maxLen `div` length x + 1) x
                      maxLen = maximum $ map length xs
                  in  sortBy (flip $ comparing pad) xs

      maxcat = read . concat . sorted . map show

Output:

[998764543431,6054854654]

Since repeating numerical string "1234" is the same as taking all the digits of 1234/9999 after the decimal point, the following does essentially the same as above:

import Data.List (sortBy)
import Data.Ord (comparing)
import Data.Ratio ((%))

nines = iterate ((+9).(*10)) 9

maxcat = foldl (\a (n,d)->a * (1 + d) + n) 0 .
    sortBy (flip $ comparing $ uncurry (%)) .
    map (\a->(a, head $ dropWhile (<a) nines))

main = mapM_ (print.maxcat) [[1,34,3,98,9,76,45,4], [54,546,548,60]]

Sort on comparison of concatenated ints method

import Data.List (sortBy)

main = print (map maxcat [[1,34,3,98,9,76,45,4], [54,546,548,60]] :: [Integer])
    where sorted = sortBy (\a b -> compare (b++a) (a++b))
          maxcat = read . concat . sorted . map show

Output as above.

Try all permutations method

import Data.List (permutations)

main :: IO ()
main =
  print
    (maxcat <$> [[1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60]] :: [Integer])
  where
    maxcat = read . maximum . fmap (concatMap show) . permutations

Output as above.

Icon and Unicon

This solution only works in Unicon as it uses a Heap class to do the heavy lifting.

import Collections    # For the Heap (dense priority queue) class

procedure main(a)
    write(lici(a))
end

procedure lici(a)
    every (result := "") ||:= Heap(a,,cmp).gen()
    return result
end

procedure cmp(a,b)
   return (a||b) > (b||a)
end

Sample runs:

->lici 1 34 3 98 9 76 45 4
998764543431
->lici 54 546 548 60
6054854654
->

J

Here we use the "pad the integers to the same size by repeating the digits then sort using these repeated integers as a sort key" approach: Solution:

maxlen=: [: >./ #&>
maxnum=: (0 ". ;)@(\: maxlen $&> ])@(8!:0)

Usage:

   maxnum&> 1 34 3 98 9 76 45 4 ; 54 546 548 60
998764543431 6054854654

Java

Works with: Java version 1.5+

This example sets up a comparator to order the numbers using Collections.sort as described in method #3 (padding and reverse sorting). It was also necessary to make a join method to meet the output requirements.

import java.util.*;

public class IntConcat {
 
    private static Comparator<Integer> sorter = new Comparator<Integer>(){
        @Override
        public int compare(Integer o1, Integer o2){
            String o1s = o1.toString();
            String o2s = o2.toString();
            
            if(o1s.length() == o2s.length()){
                return o2s.compareTo(o1s);
            }

            int mlen = Math.max(o1s.length(), o2s.length());
            while(o1s.length() < mlen * 2) o1s += o1s;
            while(o2s.length() < mlen * 2) o2s += o2s;
            
            return o2s.compareTo(o1s);
        }
    };
    
    public static String join(List<?> things){
        String output = "";
        for(Object obj:things){
            output += obj;
        }
        return output;
    }
    
    public static void main(String[] args){
        List<Integer> ints1 = new ArrayList<Integer>(Arrays.asList(1, 34, 3, 98, 9, 76, 45, 4));
        
        Collections.sort(ints1, sorter);
        System.out.println(join(ints1));
        
        List<Integer> ints2 = new ArrayList<Integer>(Arrays.asList(54, 546, 548, 60));
        
        Collections.sort(ints2, sorter);
        System.out.println(join(ints2));
    }
}

Works with: Java version 1.8+

import java.util.Comparator;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public interface IntConcat {
  public static Comparator<Integer> SORTER = (o1, o2) -> {
    String o1s = o1.toString();
    String o2s = o2.toString();
    
    if (o1s.length() == o2s.length()) {
      return o2s.compareTo(o1s);
    }
    
    int mlen = Math.max(o1s.length(), o2s.length());
    while (o1s.length() < mlen * 2) {
      o1s += o1s;
    }
    while (o2s.length() < mlen * 2) {
      o2s += o2s;
    }
    
    return o2s.compareTo(o1s);
  };

  public static void main(String[] args) {
    Stream<Integer> ints1 = Stream.of(
      1, 34, 3, 98, 9, 76, 45, 4
    );

    System.out.println(ints1
      .parallel()
      .sorted(SORTER)
      .map(String::valueOf)
      .collect(Collectors.joining())
    );

    Stream<Integer> ints2 = Stream.of(
      54, 546, 548, 60
    );

    System.out.println(ints2
      .parallel()
      .sorted(SORTER)
      .map(String::valueOf)
      .collect(Collectors.joining())
    );
  }
}

Output:

998764543431
6054854654

JavaScript

ES5

 (function () {
     'use strict';

     // maxCombine :: [Int] -> Int
     function maxCombine(xs) {
         return parseInt(
             xs.sort(
                 function (x, y) {
                     var a = x.toString(),
                         b = y.toString(),
                         ab = parseInt(a + b),
                         ba = parseInt(b + a);

                     return ab > ba ? -1 : (ab < ba ? 1 : 0);
                 }
             )
             .join(''), 10
         );
     }

     return [
        [1, 34, 3, 98, 9, 76, 45, 4],
        [54, 546, 548, 60]
     ].map(maxCombine);

 })();

Output:

[998764543431, 6054854654]

ES6

var maxCombine = (a) => +(a.sort((x, y) => +("" + y + x) - +("" + x + y)).join(''));

// test & output
console.log([
  [1, 34, 3, 98, 9, 76, 45, 4],
  [54, 546, 548, 60]
].map(maxCombine));

jq

Works with: jq version 1.4

Padding

For jq versions greater than 1.4, it may be necessary to change "sort_by" to "sort".

def largest_int:

  def pad(n):  . + (n - length) * .[length-1:];

  map(tostring)
  | (map(length) | max) as $max
  | map([., pad($max)]) 
  | sort_by( .[1] )
  | map( .[0] ) | reverse | join("") ;

# Examples:
([1, 34, 3, 98, 9, 76, 45, 4],
 [54, 546, 548, 60])  | largest_int

Output:

$ /usr/local/bin/jq -n -M -r -f Largest_int_from_concatenated_ints.jq
998764543431
6054854654

Custom Sort

The following uses quicksort/1:

def largest_int:
  map(tostring)
  | quicksort( .[0] + .[1] < .[1] + .[0] )
  | reverse | join("") ;

Julia

Perhaps algorithm 3 is more efficient, but algorithm 2 is decent and very easy to implement in Julia. So this solution uses algorithm 2.

function maxconcat(arr::Vector{<:Integer})
    b = sort(string.(arr); lt=(x, y) -> x * y < y * x, rev=true) |> join
    return try parse(Int, b) catch parse(BigInt, b) end
end

tests = ([1, 34, 3, 98, 9, 76, 45, 4],
         [54, 546, 548, 60],
         [1, 34, 3, 98, 9, 76, 45, 4, 54, 546, 548, 60])

for arr in tests
    println("Max concatenating in $arr:\n -> ", maxconcat(arr))
end

Output:

Max concatenating in [1, 34, 3, 98, 9, 76, 45, 4]:
 -> 998764543431
Max concatenating in [54, 546, 548, 60]:
 -> 6054854654
Max concatenating in [1, 34, 3, 98, 9, 76, 45, 4, 54, 546, 548, 60]:
 -> 9987660548546544543431

Kotlin

Translation of: C#

import kotlin.Comparator

fun main(args: Array<String>) {
    val comparator = Comparator<Int> { x, y -> "$x$y".compareTo("$y$x") }

    fun findLargestSequence(array: IntArray): String {
        return array.sortedWith(comparator.reversed()).joinToString("") { it.toString() }
    }

    for (array in listOf(
        intArrayOf(1, 34, 3, 98, 9, 76, 45, 4),
        intArrayOf(54, 546, 548, 60),
    )) {
        println("%s -> %s".format(array.contentToString(), findLargestSequence(array)))
    }
}

Output:

[1, 34, 3, 98, 9, 76, 45, 4] -> 998764543431
[54, 546, 548, 60] -> 6054854654

Lua

Translation of: Python

function icsort(numbers)
	table.sort(numbers,function(x,y) return (x..y) > (y..x) end)
	return numbers
end

for _,numbers in pairs({{1, 34, 3, 98, 9, 76, 45, 4}, {54, 546, 548, 60}}) do	
	print(('Numbers: {%s}\n  Largest integer: %s'):format(
		table.concat(numbers,","),table.concat(icsort(numbers))
	))
end

Output:

Numbers: {1,34,3,98,9,76,45,4}
  Largest integer: 998764543431
Numbers: {54,546,548,60}
  Largest integer: 6054854654

Mathematica/Wolfram Language

makeLargestInt[list_] := Module[{sortedlist},
  sortedlist = Sort[list, Order[ToString[#1] <> ToString[#2], ToString[#2] <> ToString[#1]] < 0 &];
  Map[ToString, sortedlist] // StringJoin // FromDigits
  ]
(* testing with two examples *)
makeLargestInt[{1, 34, 3, 98, 9, 76, 45, 4}]
makeLargestInt[{54, 546, 548, 60}]

Output:

998764543431
6054854654

Maxima

/* Function that decompose a number into a list of its digits using conversions between numbers and strings */
decompose_n_s(n):=block(
string(n),
charlist(%%),
map(eval_string,%%))$

/* Function that orders the list obtained by decompose_n_ according to ordergreat and then orders the result to reached what is needed to solve the problem */
largest_from_list(lst):=(
    sort(map(decompose_n_s,lst),ordergreatp),
    sort(%%,lambda([a,b],if last(a)>last(b) then rest(b,-1)=a else rest(a,-1)=b)),
    map(string,flatten(%%)),
    simplode(%%),
    eval_string(%%));

/* Test cases */
test1: [1, 34, 3, 98, 9, 76, 45, 4]$
test2: [54, 546, 548, 60]$
largest_from_list(test1);
largest_from_list(test2);

Output:

998764543431
6054854654

min

Works with: min version 0.19.6

(quote cons "" join) :s+
('string map (over over swap s+ 's+ dip <) sort "" join int) :fn

(1 34 3 98 9 76 45 4) fn puts!
(54 546 548 60) fn puts!

Output:

998764543431
6054854654

NetRexx

/* NetRexx */
options replace format comments java crossref symbols nobinary

runSample(arg)
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method largestInt(il) public static
  ri = ''
  wa = ''
  -- put the list into an indexed string
  wa[0] = il.words
  loop ww = 1 to wa[0]
    wa[ww] = il.word(ww)
    end ww

  -- order the list
  loop wx = 1 to wa[0] - 1
    loop wy = wx + 1 to wa[0]
      xx = wa[wx]
      yy = wa[wy]
      xy = xx || yy
      yx = yy || xx
      if xy < yx then do
        -- swap xx and yy
        wa[wx] = yy
        wa[wy] = xx
        end
      end wy
    end wx

  -- rebuild list from indexed string
  loop ww = 1 to wa[0]
    ri = ri wa[ww]
    end ww
  return ri.space(0) -- concatenate the list elements into a single numeric

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
  ints = [ -
    '1 34 3 98 9 76 45 4', -
    '54 546 548 60' -
    ]
  loop il over ints
    say largestInt(il).right(20) ':' il.space(1, ',')
    end il
  return

Output:

        998764543431 : 1,34,3,98,9,76,45,4
          6054854654 : 54,546,548,60

Nim

import algorithm, sequtils, strutils, sugar

proc maxNum(x: seq[int]): string =
  var c = x.mapIt($it)
  c.sort((x, y) => cmp(y&x, x&y))
  c.join()

echo maxNum(@[1, 34, 3, 98, 9, 76, 45, 4])
echo maxNum(@[54, 546, 548, 60])

Output:

998764543431
6054854654

OCaml

let myCompare a b = compare (b ^ a) (a ^ b)
let icsort nums = String.concat "" (List.sort myCompare (List.map string_of_int nums))

testing

# icsort [1;34;3;98;9;76;45;4];;  
- : string = "998764543431"
# icsort [54;546;548;60];;
- : string = "6054854654"

Oforth

: largestInt  map(#asString) sortWith(#[ 2dup + -rot swap + > ]) sum asInteger ;

Output:

[ [1, 34, 3, 98, 9, 76, 45, 4], [54, 546, 548, 60] ] map(#largestInt) .
[998764543431, 6054854654]

PARI/GP

Sorts then joins. Most of the noise comes from converting a vector of integers into a concatenated integer: eval(concat(apply(n->Str(n),v))). Note that the short form eval(concat(apply(Str,v))) is not valid here because Str is variadic.

large(v)=eval(concat(apply(n->Str(n),vecsort(v,(x,y)->eval(Str(y,x,"-",x,y))))));
large([1, 34, 3, 98, 9, 76, 45, 4])
large([54, 546, 548, 60])

Output:

%1 = 998764543431
%2 = 6054854654

Pascal

tested with freepascal.Used a more extreme example 3.

algorithm 3

const
  base    = 10;
  MaxDigitCnt = 11;
  source1 : array[0..7] of integer = (1, 34, 3, 98, 9, 76, 45, 4);
  source2 : array[0..3] of integer = (54,546,548,60);
  source3 : array[0..3] of integer = (60, 54,545454546,0);

type
  tdata = record
            datOrg,
            datMod : LongWord;
            datStrOrg       : string[MaxDigitCnt];
          end;
  tArrData = array of tData;

procedure DigitCount(var n: tdata);
begin
  with n do
    //InttoStr is very fast
    str(datOrg,datStrOrg);

end;

procedure InsertData(var n: tdata;data:LongWord);
begin
  n.datOrg := data;
  DigitCount(n);
end;

function FindMaxLen(const ArrData:tArrData): LongWord;
var
  cnt : longInt;
  res,t : LongWord;
begin
  res := 0;// 1 is minimum
  for cnt :=  High(ArrData) downto Low(ArrData) do
  begin
    t := length(ArrData[cnt].datStrOrg);
    IF res < t then
      res := t;
  end;
  FindMaxLen := res;
end;

procedure ExtendCount(var ArrData:tArrData;newLen: integer);
var
  cnt,
  i,k : integer;
begin
  For cnt := High(ArrData) downto Low(ArrData) do
    with ArrData[cnt] do
    begin
      datMod := datOrg;
      i := newlen-length(datStrOrg);
      k := 1;
      while i > 0 do
      begin
        datMod := datMod *Base+Ord(datStrOrg[k])-Ord('0');
        inc(k);
        IF k >length(datStrOrg) then
          k := 1;
        dec(i);
      end;
    end;
end;

procedure SortArrData(var ArrData:tArrData);
var
  i,
  j,idx : integer;
  tmpData : tData;
begin
  For i := High(ArrData) downto Low(ArrData)+1 do
  begin
    idx := i;
    j := i-1;
    For j := j downto Low(ArrData) do
      IF ArrData[idx].datMod < ArrData[j].datMod then
         idx := j;
    IF idx <> i then
    begin
      tmpData     := ArrData[idx];
      ArrData[idx]:= ArrData[i];
      ArrData[i]  := tmpData;
    end;
  end;
end;

procedure ArrDataOutput(const ArrData:tArrData);
var
  i,l : integer;
  s : AnsiString;
begin
{ the easy way
  For i := High(ArrData) downto Low(ArrData) do
    write(ArrData[i].datStrOrg);
  writeln;
  *}
  l := 0;
  For i := High(ArrData) downto Low(ArrData) do
    inc(l,length(ArrData[i].datStrOrg));
  setlength(s,l);
  l:= 1;
  For i := High(ArrData) downto Low(ArrData) do
    with ArrData[i] do
    begin
      move(datStrOrg[1],s[l],length(datStrOrg));
      inc(l,length(datStrOrg));
    end;
  writeln(s);
end;

procedure HighestInt(var  ArrData:tArrData);
begin
  ExtendCount(ArrData,FindMaxLen(ArrData));
  SortArrData(ArrData);
  ArrDataOutput(ArrData);
end;

var
  i : integer;
  tmpData : tArrData;
begin
  // Source1
  setlength(tmpData,length(source1));
  For i := low(tmpData) to high(tmpData) do
    InsertData(tmpData[i],source1[i]);
  HighestInt(tmpData);
  // Source2
  setlength(tmpData,length(source2));
  For i := low(tmpData) to high(tmpData) do
    InsertData(tmpData[i],source2[i]);
  HighestInt(tmpData);
  // Source3
  setlength(tmpData,length(source3));
  For i := low(tmpData) to high(tmpData) do
    InsertData(tmpData[i],source3[i]);
  HighestInt(tmpData);
end.

Output:

998764543431
6054854654
60545454546540

Inspired by Haskell

generate the repetition by dividing /(10^CountDigits-1) http://rosettacode.org/wiki/Largest_int_from_concatenated_ints#Compare_repeated_string_method

const
  base    = 10;
  MaxDigitCnt = 11;
  source1 : array[0..7] of LongInt = (10 , 34, 3, 98, 9, 76, 45, 4);
  source2 : array[0..3] of LongInt = (54,546,548,60);
  source3 : array[0..3] of LongInt = (0,2121212122,21,60);

type
  tdata = record
            datMod : double;
            datOrg : LongInt;
//InttoStr is very fast and the string is always needed
            datStrOrg       : string[MaxDigitCnt];
          end;
  tArrData = array of tData;

procedure InsertData(var n: tdata;data:LongWord);
begin
  with n do
  begin
    datOrg := data;
    str(datOrg,datStrOrg);
  end;
end;

function FindMaxLen(const ArrData:tArrData): LongWord;
var
  cnt : longInt;
  res,t : LongWord;
begin
  res := 0;// 1 is minimum
  for cnt :=  High(ArrData) downto Low(ArrData) do
  begin
    t := length(ArrData[cnt].datStrOrg);
    IF res < t then
      res := t;
  end;
  FindMaxLen := res;
end;

procedure ExtendData(var ArrData:tArrData;newLen: integer);
var
  cnt,
  i : integer;
begin
  For cnt := High(ArrData) downto Low(ArrData) do
    with ArrData[cnt] do
    begin
      //generating 10^length(datStrOrg)
      datMod := 1;
      i := length(datStrOrg);
      // i always >= 1
      repeat
        datMod := base*datMod;
        dec(i);
      until i <= 0;
//      1/(datMod-1.0) = 1/(9...9)
      datMod := datOrg/(datMod-1.0)+datOrg;
      i := newlen-length(datStrOrg);
      For i := i downto 1 do
        datMod := datMod*Base;
    end;
end;

procedure SortArrData(var ArrData:tArrData);
//selection sort
var
  i,
  j,idx : integer;
  tmpData : tData;
begin
  For i := High(ArrData) downto Low(ArrData)+1 do
  begin
    idx := i;
    j := i-1;
    //select max
    For j := j downto Low(ArrData) do
      IF ArrData[idx].datMod < ArrData[j].datMod then
         idx := j;
    //finally swap
    IF idx <> i then
    begin
      tmpData     := ArrData[idx];
      ArrData[idx]:= ArrData[i];
      ArrData[i]  := tmpData;
    end;
  end;
end;

procedure ArrDataOutput(const ArrData:tArrData);
var
  i : integer;
begin
{ the easy way}
  For i := High(ArrData) downto Low(ArrData) do
    write(ArrData[i].datStrOrg);
  writeln;
end;

procedure HighestInt(var  ArrData:tArrData);
begin
  ExtendData(ArrData,FindMaxLen(ArrData));
  SortArrData(ArrData);
  ArrDataOutput(ArrData);
end;

var
  i : integer;
  tmpData : tArrData;
begin
  // Source1
  setlength(tmpData,length(source1));
  For i := low(tmpData) to high(tmpData) do
    InsertData(tmpData[i],source1[i]);
  HighestInt(tmpData);
  // Source2
  setlength(tmpData,length(source2));
  For i := low(tmpData) to high(tmpData) do
    InsertData(tmpData[i],source2[i]);
  HighestInt(tmpData);
  // Source3
  setlength(tmpData,length(source3));
  For i := low(tmpData) to high(tmpData) do
    InsertData(tmpData[i],source3[i]);
  HighestInt(tmpData);
end.

Output:

9987645434310
6054854654
602121212122210>

Perl

sub maxnum {
    join '', sort { "$b$a" cmp "$a$b" } @_
}

print maxnum(1, 34, 3, 98, 9, 76, 45, 4), "\n";
print maxnum(54, 546, 548, 60), "\n";

Output:

998764543431
6054854654

Phix

with javascript_semantics
function catcmp(string a, string b)
    return compare(b&a,a&b)
end function
 
function method2(sequence s)
    return join(custom_sort(catcmp,apply(s,sprint)),"")
end function
 
?method2({1,34,3,98,9,76,45,4})
?method2({54,546,548,60})

Output:

"998764543431"
"6054854654"

PHP

function maxnum($nums) {
    usort($nums,  function ($x, $y) { return strcmp("$y$x", "$x$y"); });
    return implode('', $nums);
}

echo maxnum(array(1, 34, 3, 98, 9, 76, 45, 4)), "\n";
echo maxnum(array(54, 546, 548, 60)), "\n";

Output:

998764543431
6054854654

Picat

On the simpler cases, four methods are tested: 2 using permutations and 2 with different sorting methods.

permutation/2

s_perm1(L, Num) => 
  permutation(L,P),
  Num = [I.to_string() : I in P].flatten().to_integer().

Using permutations/1

s_perm2(L, Num) => 
  Perms = permutations(L),
  Num = max([  [I.to_string() : I in P].flatten().to_integer() : P in Perms]).

Sort on concatenated numbers

s_sort_conc(L,Num) =>
  Num = [to_string(I) : I in qsort(L,f3)].join('').to_integer().

% sort function for s_sort_conc/2
f3(N1,N2) => 
  N1S = N1.to_string(),
  N2S = N2.to_string(),
  (N1S ++ N2S).to_integer() >= (N2S ++ N1S).to_integer().

% qsort(List, SortFunction)
% returns a sorted list according to the sort function SortFunction.
qsort([],_F)    = [].
qsort([H|T],F) = qsort([E : E in T, call(F,E,H)], F) 
                 ++ [H] ++
                 qsort([E : E in T, not call(F,E,H)],F).

Extend each element to the largest length

s_extend(L,Num) =>
  LS =  [I.to_string() : I in L],
  MaxLen = 2*max([I.length : I in LS]),
  L2 = [],
  foreach(I in LS)
    I2 = I,
    % extend to a larger length
    while(I2.length < MaxLen)
     I2 := I2 ++ I
    end,
    % keep info of the original number
    L2 := L2 ++ [[I2,I]]
  end,
  Num = [I[2] : I in qsort(L2,f4)].join('').to_integer().

% sort function for s_extend/2
f4(N1,N2) => N1[1].to_integer() >= N2[1].to_integer().

Test

import util.

go =>
  Ls = [[1, 34, 3, 98, 9, 76, 45, 4],
        [54, 546, 548, 60],
        [97, 9, 13, 979],
        [9, 1, 95, 17, 5]
        ],
  foreach(L in Ls)
    test(L)
  end,
  nl.

  
% Test all implementations
test(L) =>
  println(l=L),

  maxof_inc(s_perm1(L,Num1), Num1),
  println(s_perm1=Num1),

  s_perm2(L,Num2),
  println(s_perm2=Num2),

  s_sort_conc(L,Num3),
  println(s_sort_conc=Num3),

  s_extend(L,Num4),
  println(s_extent=Num4),
  nl.

Output:

l = [1,34,3,98,9,76,45,4]
s_perm1     = 998764543431
s_perm2     = 998764543431
s_sort_conc = 998764543431
s_extend    = 998764543431

l = [54,546,548,60]
s_perm1     = 6054854654
s_perm2     = 6054854654
s_sort_conc = 6054854654
s_extend    = 6054854654

l = [97,9,13,979]
s_perm1     = 99799713
s_perm2     = 99799713
s_sort_conc = 99799713
s_extend    = 99799713

l = [9,1,95,17,5]
s_perm1     = 9955171
s_perm2     = 9955171
s_sort_conc = 9955171
s_extend    = 9955171

Testing larger instance

Test a larger instance: 2000 random numbers between 1 and 100; about 5800 digits. The two permutation variants (s_perm1 and s_perm2) takes too long on larger N, say N > 9.

go2 =>
  garbage_collect(100_000_000),
  _ = random2(),
  N = 2000,
  println(nums=N),
  L = [random(1,1000) : _ in 1..N],
  S = join([I.to_string : I in L],''),
  println(str_len=S.len),  

  nl,
  println("s_sort_conc:"),
  time(s_sort_conc(L,_Num3)),

  println("s_extend:"),
  time(s_extend(L,_Num4)),

  nl.

Output:

nums = 2000
str_len = 5805

s_sort_conc:

CPU time 0.54 seconds.

s_extend:

CPU time 0.526 seconds.

PicoLisp

Here are solutions for all three algorithms.

The third solution actually avoids padding the numbers, by converting them into circular lists and comparing these. As a drawback, however, this works only for unique lists (as the comparison of identical numbers would not terminate), so a better solution might involve additional checks.

(load "@lib/simul.l")  # For 'permute'

Algorithm 1

(for L '((1 34 3 98 9 76 45 4) (54 546 548 60))
   (prinl (maxi format (permute L))) )

Algorithm 2

(for L '((1 34 3 98 9 76 45 4) (54 546 548 60))
   (prinl
      (sort L
         '((A B)
            (>
               (format (pack A B))
               (format (pack B A)) ) ) ) ) )

Algorithm 3

(for L '((1 34 3 98 9 76 45 4) (54 546 548 60))
   (prinl
      (flip
         (by '((N) (apply circ (chop N))) sort L) ) ) )

Output:

in all three cases

998764543431
6054854654

PL/I

/* Largest catenation of integers            16 October 2013 */
/* Sort using method 2, comparing pairs of adjacent integers. */

Largest: procedure options (main);
   declare s(*) char (20) varying controlled, n fixed binary;
   get (n);
   allocate s(n);
   get list (s);
   s = trim(s);
   put skip edit (s) (a, x(1));
   put skip list ('Largest integer=', Largest_integer());

largest_integer: procedure () returns (char(100) varying);
   declare sorted bit (1);
   declare (true value ('1'b), false value ('0'b)) bit (1);
   declare i fixed binary;
   declare temp character(20) varying;

   do until (sorted);
      sorted = true;
      do i = 1 to n-1;
         if char(s(i)) || char(s(i+1)) < char(s(i+1)) || char(s(i)) then
            do;
               temp = s(i); s(i) = s(i+1); s(i+1) = temp; sorted = false;
            end;
      end;
   end;
   return (string(s));
end largest_integer;
end Largest;

54 546 548 60
Largest integer=        6054854654 

1 34 3 98 9 76 45 4
Largest integer=        998764543431 

PowerShell

Works with: PowerShell version 2

Using algorithm 3

Function Get-LargestConcatenation ( [int[]]$Integers )
    {
    #  Get the length of the largest integer
    $Length = ( $Integers | Sort -Descending | Select -First 1 ).ToString().Length
 
    #  Convert to an array of strings,
    #  sort by each number repeated Length times and truncated to Length,
    #  and concatenate (join)
    $Concat = ( [string[]]$Integers | Sort { ( $_ * $Length ).Substring( 0, $Length ) } -Descending ) -join ''
 
    #  Convert to integer (upsizing type if needed)
    try           { $Integer = [ int32]$Concat }
    catch { try   { $Integer = [ int64]$Concat }
            catch { $Integer = [bigint]$Concat } }
 
    return $Integer
    }

Get-LargestConcatenation 1, 34, 3, 98, 9, 76, 45, 4
Get-LargestConcatenation 54, 546, 548, 60
Get-LargestConcatenation 54, 546, 548, 60, 54, 546, 548, 60

Output:

998764543431
6054854654
60605485485465465454

Prolog

Works with SWI-Prolog 6.5.3.

All permutations method

largest_int_v1(In, Out) :-
	maplist(name, In, LC),
	aggregate(max(V), get_int(LC, V), Out).


get_int(LC, V) :-
	permutation(LC, P),
	append(P, LV),
	name(V, LV).

Output:

 ?- largest_int_v1([1, 34, 3, 98, 9, 76, 45, 4], Out).
Out = 998764543431.

 ?- largest_int_v1([54, 546, 548, 60], Out).
Out = 6054854654.

Method 2

largest_int_v2(In, Out) :-
	maplist(name, In, LC),
	predsort(my_sort,LC, LCS),
	append(LCS, LC1),
	name(Out, LC1).


my_sort(R, L1, L2) :-
	append(L1, L2, V1), name(I1, V1),
	append(L2, L1, V2), name(I2, V2),
	(   I1 < I2, R = >; I1 = I2, R = '='; R = <).



% particular case  95 958
my_sort(>, [H1], [H1,  H2 | _]) :-
	H1 > H2.

my_sort(<, [H1], [H1, H2 | _]) :-
	H1 < H2.

my_sort(R, [H1], [H1, H1 | T]) :-
	my_sort(R, [H1], [H1 | T]).



% particular case  958 95
my_sort(>, [H1,  H2 | _], [H1]) :-
	H1 > H2.

my_sort(<, [H1,  H2 | _], [H1]) :-
	H1 < H2.

my_sort(R, [H1,  H1 | T], [H1]) :-
	my_sort(R, [H1 | T], [H1]) .

Output:

 ?- largest_int_v2([1, 34, 3, 98, 9, 76, 45, 4], Out).
Out = 998764543431 .

 ?- largest_int_v2([54, 546, 548, 60], Out).
Out = 5486054654 .

Python

Python: Sort on comparison of concatenated ints method

This also shows one of the few times where cmp= is better than key= on sorted()

try:
    cmp     # Python 2 OK or NameError in Python 3
    def maxnum(x):
        return ''.join(sorted((str(n) for n in x),
                              cmp=lambda x,y:cmp(y+x, x+y)))
except NameError:
    # Python 3
    from functools import cmp_to_key
    def cmp(x, y):
        return -1 if x<y else ( 0 if x==y else 1)
    def maxnum(x):
        return ''.join(sorted((str(n) for n in x),
                              key=cmp_to_key(lambda x,y:cmp(y+x, x+y))))

for numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
    print('Numbers: %r\n  Largest integer: %15s' % (numbers, maxnum(numbers)))

Output:

Numbers: (1, 34, 3, 98, 9, 76, 45, 4)
  Largest integer:    998764543431
Numbers: (54, 546, 548, 60)
  Largest integer:      6054854654

Python: Compare repeated string method

def maxnum(x):
    maxlen = len(str(max(x)))
    return ''.join(sorted((str(v) for v in x), reverse=True,
                          key=lambda i: i*(maxlen * 2 // len(i))))

for numbers in [(212, 21221), (1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
    print('Numbers: %r\n  Largest integer: %15s' % (numbers, maxnum(numbers)))

Output:

Numbers: (212, 21221)
  Largest integer:        21221221
Numbers: (1, 34, 3, 98, 9, 76, 45, 4)
  Largest integer:    998764543431
Numbers: (54, 546, 548, 60)
  Largest integer:      6054854654

Works with: Python version 2.6+

from fractions import Fraction
from math import log10

def maxnum(x):
    return ''.join(str(n) for n in sorted(x, reverse=True,
                          key=lambda i: Fraction(i, 10**(int(log10(i))+1)-1)))

for numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
    print('Numbers: %r\n  Largest integer: %15s' % (numbers, maxnum(numbers)))

Output as first Python example, above.

Python: Try all permutations method

from itertools import permutations
def maxnum(x):
    return max(int(''.join(n) for n in permutations(str(i) for i in x)))

for numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:
    print('Numbers: %r\n  Largest integer: %15s' % (numbers, maxnum(numbers)))

Output as above.

Quackery

With a string of space separated sequences of digits

[ sortwith
    [ 2dup swap join
      dip join $< ]
  [] swap witheach join ] is largest-int ( [ --> $ )

$ '1 34 3 98 9 76 45 4' nest$ largest-int echo$ cr
$ '54 546 548 60' nest$ largest-int echo$

Output:

998764543431
6054854654

With a nest of numbers

  [ number$ dip number$ join $->n drop ] is conc   ( n n --> n )

  [ 2dup conc unrot swap conc < ]        is conc>  ( n n --> b )

  [ sortwith conc>
    $ "" swap 
    witheach [ number$ join ]
    $->n drop ]                          is task (   [ --> n )

  ' [ [ 1 34 3 98 9 76 45 4 ] 
      [ 54 546 548 60 ] ] 

  witheach [ task echo cr ]

Output:

998764543431
6054854654

R

Largest_int_from_concat_ints <- function(vec){
  
  #recursive function for computing all permutations
  perm <- function(vec) {  
    n <- length(vec)
    if (n == 1) 
      return(vec)
    else {
      x <- NULL
      for (i in 1:n){ 
        x <- rbind(x, cbind(vec[i], perm(vec[-i])))
      }
      return(x)
    }
  }
  
  permutations <- perm(vec)
  concat <- as.numeric(apply(permutations, 1, paste, collapse = ""))
  return(max(concat))
}

#Verify
Largest_int_from_concat_ints(c(54, 546, 548, 60))
Largest_int_from_concat_ints(c(1, 34, 3, 98, 9, 76, 45, 4))
Largest_int_from_concat_ints(c(93, 4, 89, 21, 73))

Output:

[1] 6054854654
[1] 998764543431
[1] 938973421

Racket

#lang racket
(define (largest-int ns)
  (string->number (apply ~a (sort ns (λ(x y) (string>? (~a x y) (~a y x)))))))
(map largest-int '((1 34 3 98 9 76 45 4) (54 546 548 60)))
;; -> '(998764543431 6054854654)

Raku

(formerly Perl 6)

sub maxnum(*@x) {
    [~] @x.sort: -> $a, $b { $b ~ $a leg $a ~ $b }
}

say maxnum <1 34 3 98 9 76 45 4>;
say maxnum <54 546 548 60>;

Output:

998764543431
6054854654

Red

Red []

foreach seq [[1 34 3 98 9 76 45 4] [54 546 548 60]] [ 
  print rejoin sort/compare seq function [a b] [ (rejoin [a b]) > rejoin [b a] ] 
]

Output:

998764543431
6054854654

REXX

The algorithm used is based on exact comparisons (left to right)   with   right digit fill   of the   left digit.
This allows the integers to be of any size.

This REXX version works with any size integer   (negative, zero, positive),   and does some basic error checking to
verify that the numbers are indeed integers   (and it also normalizes the integers).

The absolute value is used for negative numbers.   No sorting of the numbers is required for the 1^st two examples.

simple integers

/*REXX program constructs the largest integer  from an integer list using concatenation.*/
@.=.;     @.1 =   1   34    3  98  9  76  45  4  /*the  1st  integer list to be used.   */
          @.2 =  54  546  548  60                /* "   2nd     "      "   "  "   "     */
          @.3 =   4   45   54   5                /* "   3rd     "      "   "  "   "     */
w=0                                              /* [↓]   process all the integer lists.*/
    do j=1  while @.j\==.;         z= space(@.j) /*keep truckin' until lists exhausted. */
    w=max(w, length(z) );          $=            /*obtain maximum width to align output.*/
        do  while z\='';  idx= 1;  big= norm(1)  /*keep examining the list  until  done.*/
          do k=2  to  words(z);    #= norm(k)    /*obtain an a number from the list.    */
          L= max(length(big), length(#) )        /*get the maximum length of the integer*/
          if left(#, L, left(#, 1) )   <<=   left(big, L, left(big, 1) )    then iterate
          big= #;                  idx= k        /*we found a new biggie (and the index)*/
          end   /*k*/                            /* [↑]  find max concatenated integer. */
        z= delword(z, idx, 1)                    /*delete this maximum integer from list*/
        $= $  ||  big                            /*append   "     "       "    ───►  $. */
        end     /*while z*/                      /* [↑]  process all integers in a list.*/
    say 'largest concatenatated integer from '  left( space(@.j), w)    " is ─────► "    $
    end         /*j*/                            /* [↑]  process each list of integers. */
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
norm: arg i;  #= word(z, i);  er= '***error***';  if left(#, 1)=="-"  then #= substr(#, 2)
      if \datatype(#,'W')  then do; say er # "isn't an integer."; exit 13; end; return #/1

output   when using the default (internal) integer lists:

largest concatenatated integer from  1 34 3 98 9 76 45 4  is ─────►  998764543431
largest concatenatated integer from  54 546 548 60        is ─────►  6054854654
largest concatenatated integer from  4 45 54 5            is ─────►  554454

exponentiated integers

In REXX, a number such as   6.6e77   would be considered an integer   if   the (current)   numeric digits   is
large enough to express that number as an integer without the exponent.

The default for REXX is   9   decimal digits,   but the   norm   function automatically uses enough decimal digits to
express the number as an integer.

This REXX version can handle any sized integer   (most REXXes can handle up to around eight million decimal
digits,   but displaying the result would be problematic for results wider than the display area).

/*REXX program constructs the largest integer  from an integer list using concatenation.*/
@.=.;     @.1 =   1   34    3  98  9  76  45  4  /*the  1st  integer list to be used.   */
          @.2 =  54  546  548  60                /* "   2nd     "      "   "  "   "     */
          @.3 =   4   45   54   5                /* "   3rd     "      "   "  "   "     */
          @.4 =   4   45   54   5  6.6e77        /* "   4th     "      "   "  "   "     */
w= 0                                             /* [↓]   process all the integer lists.*/
    do j=1  while @.j\==.;        z= space(@.j)  /*keep truckin' until lists exhausted. */
    w=max(w, length(z) );         $=             /*obtain maximum width to align output.*/
        do while z\='';  idx=1;   big= norm(1)   /*keep examining the list  until  done.*/
          do k=2  to  words(z);   #= norm(k)     /*obtain an a number from the list.    */
          L= max(length(big), length(#) )        /*get the maximum length of the integer*/
          if left(#, L, left(#, 1) )   <<=   left(big, L, left(big, 1) )    then iterate
          big=#;                  idx= k         /*we found a new biggie (and the index)*/
          end   /*k*/                            /* [↑]  find max concatenated integer. */
        z= delword(z, idx, 1)                    /*delete this maximum integer from list*/
        $= $  ||  big                            /*append   "     "       "    ───►  $. */
        end     /*while z*/                      /* [↑]  process all integers in a list.*/
    say 'largest concatenatated integer from '    left( space(@.j), w)       " is "      $
    end         /*j*/                            /* [↑]  process each list of integers. */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
norm: arg i;  #= word(z, i);  er= '***error***';  if left(#, 1)=="-"  then #= substr(#, 2)
      if \datatype(#, 'N')  then signal er13                      /*go and tell err msg.*/
                            else #= # / 1                         /*a #, so normalize it*/
      if pos('E',#)>0  then do; parse var # mant "E" pow          /*Has exponent? Expand*/
                                numeric digits pow + length(mand) /*expand digs, adjust#*/
                            end
      if datatype(#, 'W')  then return # / 1
er13: say er  #  "isn't an integer.";             exit 13

output   when using the default (internal) integer lists:

(Output shown at three-quarter size.)

largest concatenatated integer from  1 34 3 98 9 76 45 4  is  998764543431
largest concatenatated integer from  54 546 548 60        is  6054854654
largest concatenatated integer from  4 45 54 5            is  554454
largest concatenatated integer from  4 45 54 5 6.6e77     is  660000000000000000000000000000000000000000000000000000000000000000000000000000554454

Alternate Version

Inspired by the previous versions.

/*REXX program constructs the largest integer  from an integer list using concatenation.*/
l.='';    l.1 = '1 34 3 98 9 76 45 4'   /*the  1st  integer list to be used.   */
          l.2 = '54 546 548 60'             /* "   2nd     "      "   "  "   "     */
          l.3 = ' 4  45  54  5'             /* "   3rd     "      "   "  "   "     */
          l.4 = ' 4  45  54  5  6.6e77'    /* "   4th     "      "   "  "   "     */
          l.5 = ' 3 3 .2'                  /* "   5th     "      "   "  "   "     */
/*
soll.1=998764543431
soll.2=6054854654
soll.3=554454
soll.4=660000000000000000000000000000000000000000000000000000000000000000000000000000545454
*/
l_length=0
Do li=1 By 1 While l.li<>''
  l_length=max(l_length,length(space(l.li)))
  End

Do li=1 By 1 While l.li<>''
  z=''
  Do j=1 To words(l.li)
    int=integer(word(l.li,j))
    If int='?' Then Do
      Say left(space(l.li),l_length) '-> ** invalid ** bad integer' word(l.li,j)
      Iterate li
      End
    Else
      z=z int
    End
/*Say copies(' ',l_length) '  ' soll.li */
  Say left(space(l.li),l_length) '->' largeint(l.li)
  End
Exit

integer: Procedure
Numeric Digits 1000
Parse Arg z
If Datatype(z,'W') Then
  Return z+0
Else
  Return '?'

largeint:
result=''
Do While z<>''                                 /* [?]  check the rest of the integers.*/
  big=word(z,1); index=1; LB=length(big)       /*assume that first integer is biggest.*/
  do k=2 to words(z);
    n=word(z,k)                                /*obtain an integer from the list.     */
    L=max(LB,length(n))                        /*get the maximum length of the integer*/
    if left(n,L,left(n,1))<<=left(big,L,left(big,1)) then iterate
    big=n; index=k                             /*we found a new biggie (and the index)*/
    LB=length(big)
    End   /*k*/
  z=delword(z,index,1)                         /*delete this maximum integer from list*/
  result=result||big                           /*append   "     "       "    ---?  $. */
  end     /*while z*/                          /* [?]  process all integers in a list.*/
Return result

Output:

1 34 3 98 9 76 45 4 -> 998764543431
54 546 548 60       -> 6054854654
4 45 54 5           -> 554454
4 45 54 5 6.6e77    -> 660000000000000000000000000000000000000000000000000000000000000000000000000000554454
3 3 .2              -> ** invalid ** bad integer .2

Version 4

Translation of: NetRexx

/*REXX program constructs the largest integer from an integer list using concatenation.*/
l.='';    l.1 = '1 34 3 98 9 76 45 4'           /*the  1st  integer list to be used.   */
          l.2 = '54 546 548 60'                 /* "   2nd     "      "   "  "   "     */
          l.3 = ' 4  45  54  5'                 /* "   3rd     "      "   "  "   "     */
          l.4 = ' 4  45  54  5  6.6e77'         /* "   4th     "      "   "  "   "     */
          l.5 = ' 3 3 .2'                       /* "   5th     "      "   "  "   "     */
          l.6 = ' 4  45  54  5  6.6e1001'       /* "   6th     "      "   "  "   "     */
          l.7 = ' 4.0000 45 54 5.00'            /* "   7th     "      "   "  "   "     */
          l.8 = ' 10e999999999 5'               /* "   8th     "      "   "  "   "     */
l_length=0
Do li=1 By 1 While l.li<>''
  l_length=max(l_length,length(space(l.li)))
  End

Do li=1 By 1 While l.li<>''
  z=''
  msg=''
  Do j=1 To words(l.li)
    int=integer(word(l.li,j))
    If int='?' Then Do
      Say left(space(l.li),l_length) '-> ** invalid ** bad list item:' word(l.li,j) msg
      Iterate li
      End
    Else
      z=z int
    End
  zz=largeint(z)
  If length(zz)<60 Then
    Say left(space(l.li),l_length) '->' zz
  Else
    Say left(space(l.li),l_length) '->' left(zz,5)'...'right(zz,5)
  End
Exit

integer: Procedure Expose msg
Numeric Digits 1000
Parse Arg z
If Datatype(z,'W') Then
  Return z/1
Else Do
  If Datatype(z,'NUM') Then Do
    Do i=1 To 6 Until dig>=999999999
      dig= digits()*10
      dig=min(dig,999999999)
      Numeric Digits dig
      If Datatype(z,'W') Then
        Return z/1
      End
    msg='cannot convert it to an integer'
    Return '?'
    End
  Else Do
    msg='not a number (larger than what this REXX can handle)'
    Return '?'
    End
  End

largeint: Procedure
Parse Arg list
w.0=words(list)
Do i=1 To w.0
  w.i=word(list,i)
  End
Do wx=1 To w.0-1
  Do wy=wx+1 To w.0
    xx=w.wx
    yy=w.wy
    xy=xx||yy
    yx=yy||xx
    if xy < yx then do
      /* swap xx and yy */
      w.wx = yy
      w.wy = xx
      end
    End
  End
list=''
Do ww=1 To w.0
  list=list w.ww
  End
Return space(list,0)

Output:

1 34 3 98 9 76 45 4 -> 998764543431
54 546 548 60       -> 6054854654
4 45 54 5           -> 554454
4 45 54 5 6.6e77    -> 66000...54454
3 3 .2              -> ** invalid ** bad list item: .2 cannot convert it to an integer
4 45 54 5 6.6e1001  -> 66000...54454
4.0000 45 54 5.00   -> 554454
10e999999999 5      -> ** invalid ** bad list item: 10e999999999 not a number (larger than what this REXX can handle)

Ring

nums=[1,34,3,98,9,76,45,4]
see largestInt(8) + nl
nums=[54,546,548,60]
see largestInt(4) + nl
 
func largestInt len
l = ""
sorted = false
while not sorted
      sorted=true
      for i=1 to len-1
          a=string(nums[i])
          b=string(nums[i+1])
          if a+b<b+a 
             temp = nums[i]
             nums[i] = nums[i+1]
             nums[i+1] = temp
             sorted=false ok
      next
end
for i=1 to len
    l+=string(nums[i])
next
return l

Output:

998764543431
6054854654

RPL

We use here the second algorithm, easily derived from the SORT program given in Sorting algorithms/Bubble sort.

Works with: HP version 28

≪ LIST→ → len 
   ≪ 1 len START 
        →STR len ROLL END
      len 1 FOR n 
         1 n 1 - START 
            IF DUP2 + LAST SWAP + < THEN SWAP END 
            n ROLLD 
         NEXT n ROLLD  
      -1 STEP 
      2 len START + END STR→
≫ ≫ ‘MKBIG’ STO

{212 21221} MKBIG
{1 34 3 98 9 76 45 4} MKBIG
{54 546 548 60} MKBIG

Output:

3: 21221221
2: 998764543431
1: 6054854654

Ruby

Sort on comparison of concatenated ints method

Translation of: Tcl

def icsort nums
  nums.sort { |x, y| "#{y}#{x}" <=> "#{x}#{y}" }
end

[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each do |c|
  p c # prints nicer in Ruby 1.8
  puts icsort(c).join
end

Output:

[54, 546, 548, 60]
6054854654
[1, 34, 3, 98, 9, 76, 45, 4]
998764543431

Compare repeated string method

def icsort nums
  maxlen = nums.max.to_s.length
  nums.map{ |x| x.to_s }.sort_by { |x| x * (maxlen * 2 / x.length) }.reverse
end

[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each do |c|
  p c # prints nicer in Ruby 1.8
  puts icsort(c).join
end

Output as above.

require 'rational' #Only needed in Ruby < 1.9

def icsort nums
  nums.sort_by { |i| Rational(i, 10**(Math.log10(i).to_i+1)-1) }.reverse
end

[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each do |c|
  p c # prints nicer in Ruby 1.8
  puts icsort(c).join
end

Output as above.

Run BASIC

a1$ = "1, 34, 3, 98, 9, 76, 45, 4"
a2$ = "54,546,548,60"

print "Max Num ";a1$;" = ";maxNum$(a1$)
print "Max Num ";a2$;" = ";maxNum$(a2$)

function maxNum$(a1$)
while word$(a1$,i+1,",") <> ""
 i = i + 1
 a$(i) = trim$(word$(a1$,i,","))
wend

s = 1
while s = 1
 s = 0
 for j = 1 to i -1
  if a$(j)+a$(j+1) < a$(j+1)+a$(j) then
   h$      = a$(j)
   a$(j)   = a$(j+1)
   a$(j+1) = h$
   s       = 1
  end if
 next j
wend

for j = 1 to i
 maxNum$ = maxNum$ ; a$(j)
next j
end function

Output:

Max Num 1, 34, 3, 98, 9, 76, 45, 4 = 998764543431
Max Num 54,546,548,60 = 6054854654

Rust

fn maxcat(a: &mut [u32]) {
    a.sort_by(|x, y| {
        let xy = format!("{}{}", x, y);
        let yx = format!("{}{}", y, x);
        xy.cmp(&yx).reverse()
    });
    for x in a {
        print!("{}", x);
    }
    println!();
}
 
fn main() {
    maxcat(&mut [1, 34, 3, 98, 9, 76, 45, 4]);
    maxcat(&mut [54, 546, 548, 60]);
}

Output:

998764543431
6054854654

S-lang

define catcmp(a, b)
{
   a = string(a);
   b = string(b);
   return strcmp(b+a, a+b);
}
	  
define maxcat(arr)
{
   arr = arr[array_sort(arr, &catcmp)];
   variable result = "", elem;
   foreach elem (arr)
     result += string(elem);
   return result;
}

print("max of series 1 is " + maxcat([1, 34, 3, 98, 9, 76, 45, 4]));
print("max of series 2 is " + maxcat([54, 546, 548, 60]));

Output:

"max of series 1 is 998764543431"
"max of series 2 is 6054854654"

Scala

Library: Scala

object LIFCI extends App {

  def lifci(list: List[Long]) = list.permutations.map(_.mkString).max

  println(lifci(List(1, 34, 3, 98, 9, 76, 45, 4)))
  println(lifci(List(54, 546, 548, 60)))
}

Output:

 998764543431
 6054854654

Scheme

(define (cat . nums)  (apply string-append (map number->string nums)))

(define (my-compare a b)  (string>? (cat a b) (cat b a)))

(map  (lambda (xs) (string->number (apply cat (sort xs my-compare))))
      '((1 34 3 98 9 76 45 4) (54 546 548 60)))

Output:

(998764543431 6054854654)

Sidef

Translation of: Ruby

func maxnum(nums) {
    nums.sort {|x,y|  "#{y}#{x}" <=> "#{x}#{y}" };
}

[[54, 546, 548, 60], [1, 34, 3, 98, 9, 76, 45, 4]].each { |c|
    say maxnum(c).join.to_num;
}

Output:

6054854654
998764543431

Smalltalk

Version 1) sort by padded print strings:

Works with: Smalltalk/X

#(
    (54  546  548  60)
    (1  34  3  98  9  76  45  4)
) do:[:ints |
    |resultString|

    "sort ints by padded strings (sort a copy - literals are immudatble),
     then collect their strings, then concatenate"
    resultString :=
        ((ints copy sort:[:a :b | 
                |pad|
                pad := (a integerLog10) max:(b integerLog10).
                (a printString paddedTo:pad with:$0) > (b printString paddedTo:pad with:$0)])
            collect:#printString) asStringWith:''.
    Stdout printCR: resultString.
].

Version 2) alternative: sort by concatenated pair's strings:

#(
    (54  546  548  60)
    (1  34  3  98  9  76  45  4)
) do:[:ints |
    |resultString|

    resultString :=
        ((ints copy sort:[:a :b | e'{a}{b}' > e'{b}{a}']) "(1)"
            collect:#printString) asStringWith:''.
    Stdout printCR: resultString.
].

Note ¹ replace "e'{a}{b}'" by "(a printString,b printString)" in dialects, which do not support embedded expression strings.

Version 3) no need to collect the resultString; simply print the sorted list (ok, if printing is all we want):

#(
    (54  546  548  60)
    (1  34  3  98  9  76  45  4)
) do:[:ints |
    (ints copy sort:[:a :b | e'{a}{b}' > e'{b}{a}'])
        do:[:eachNr | eachNr printOn:Stdout].
    Stdout cr.
]

Version 4) no need to generate any intermediate strings; the following will do as well:

#(
    (54  546  548  60)
    (1  34  3  98  9  76  45  4)
) do:[:ints |
    (ints copy sortByApplying:[:i | i log10 fractionPart]) reverseDo:#print.
    Stdout cr.
]

Output:

6054854654
989764543431

Tcl

proc intcatsort {nums} {
    lsort -command {apply {{x y} {expr {"$y$x" - "$x$y"}}}} $nums
}

Demonstrating:

foreach collection {
    {1 34 3 98 9 76 45 4}
    {54 546 548 60}
} {
    set sorted [intcatsort $collection]
    puts "\[$collection\] => \[$sorted\]  (concatenated: [join $sorted ""])"
}

Output:

[1 34 3 98 9 76 45 4] => [9 98 76 45 4 34 3 1]  (concatenated: 998764543431)
[54 546 548 60] => [60 548 546 54]  (concatenated: 6054854654)

Transd

#lang transd

MainModule: {
_start: (lambda
    (for ar in [[98, 76, 45, 34, 9, 4, 3, 1], [54, 546, 548, 60]] do
        (sort ar (λ l Int() r Int() (ret (> Int(String(l r)) Int(String(r l))))))
        (lout (join ar "")))
)
}

Output:

998764543431
6054854654

VBScript

Translation of: BBC BASIC

Function largestint(list)
	nums = Split(list,",")
	Do Until IsSorted = True
		IsSorted = True
		For i = 0 To UBound(nums)
			If i <> UBound(nums) Then
				a = nums(i)
				b = nums(i+1)
				If CLng(a&b) < CLng(b&a) Then
					tmpnum = nums(i)
					nums(i) = nums(i+1)
					nums(i+1) = tmpnum
					IsSorted = False
				End If
			End If
		Next
	Loop
	For j = 0 To UBound(nums)
		largestint = largestint & nums(j)
	Next
End Function

WScript.StdOut.Write largestint(WScript.Arguments(0))
WScript.StdOut.WriteLine

Output:

F:\>cscript /nologo largestint.vbs 1,34,3,98,9,76,45,4
998764543431

F:\>cscript /nologo largestint.vbs 54,546,548,60
6054854654

Vim Script

This solution is intended to be run as an Ex command within a buffer containing the integers to be processed, one per line.

%s/\(.\+\)/\1\1/ | sort! | %s/\(.\+\)\1\n/\1/

Demonstration

$ paste -s nums
1	34	3	98	9	76	45	4
$ vim -S icsort.vim nums
998764543431

Wren

Translation of: Kotlin

Library: Wren-sort

import "./sort" for Sort

var cmp = Fn.new { |x, y|
    var xy = Num.fromString(x.toString + y.toString)
    var yx = Num.fromString(y.toString + x.toString)
    return (xy - yx).sign
}

var findLargestSequence = Fn.new { |a|
    var b = Sort.merge(a, cmp)
    return b[-1..0].join()
}

var arrays = [
    [1, 34, 3, 98, 9, 76, 45, 4],
    [54, 546, 548, 60]
]
for (a in arrays) {
    System.print("%(a) -> %(findLargestSequence.call(a))")
}

Output:

[1, 34, 3, 98, 9, 76, 45, 4] -> 998764543431
[54, 546, 548, 60] -> 6054854654

zkl

fcn bigCI(ns){
   ns.apply("toString").sort(fcn(a,b){ (a+b)>(b+a) }).concat();
}

bigCI(T(1, 34, 3, 98, 9, 76, 45, 4)).println();
bigCI(T(54, 546, 548, 60)).println();

Output:

998764543431
6054854654